Merge pull request soapyigu#352 from soapyigu/DFS

Add a solution to Number of Closed Islands

Author: soapyigu

DFS/NumberClosedIslands.swift

@@ -0,0 +1,59 @@
+/**
+ * Question Link: https://leetcode.com/problems/number-of-closed-islands/
+ * Primary idea: Classic Depth-first Search, go up, down, left, right four directions. Return false only it hits the edge.
+ * 
+ * Time Complexity: O(mn), Space Complexity: O(mn)
+ *
+ */
+
+class NumberClosedIslands {
+    func closedIsland(_ grid: [[Int]]) -> Int {
+        let m = grid.count, n = grid[0].count
+        
+        var isVisited = Array(repeating: Array(repeating: false, count: n), count: m), res = 0
+        
+        for i in 0..<m {
+            for j in 0..<n {
+                guard !isVisited[i][j], grid[i][j] == 0 else {
+                    continue
+                }
+                
+                if dfs(grid, i, j, m, n, &isVisited) {
+                    res += 1
+                }
+            }
+        }
+        
+        return res
+    }
+    
+    private func dfs(_ grid: [[Int]], _ i: Int, _ j: Int, _ m: Int, _ n: Int, _ isVisited: inout [[Bool]]) -> Bool {
+        guard i >= 0 && i < m && j >= 0 && j < n else {
+            return false
+        }
+        
+        if grid[i][j] == 1 {
+            return true
+        }
+        
+        isVisited[i][j] = true
+        
+        
+        var up = true, down = true, left = true, right = true
+    
+        if i - 1 < 0 || !isVisited[i - 1][j] {
+            up = dfs(grid, i - 1, j, m, n, &isVisited)
+        }
+        if i + 1 >= m || !isVisited[i + 1][j] {
+            down = dfs(grid, i + 1, j, m, n, &isVisited)
+        }
+        if j - 1 < 0 || !isVisited[i][j - 1] {
+            left = dfs(grid, i, j - 1, m, n, &isVisited)
+        }
+        if j + 1 >= n || !isVisited[i][j + 1] {
+            right = dfs(grid, i, j + 1, m, n, &isVisited)
+        }
+          
+        return up && down && left && right
+    }
+}